Question

How many moles of \(\mathrm{HF}\left(K_{a}=6.8 \times 10^{-4}\right)\) must be present in \(0.200 \mathrm{~L}\) to form a solution with a \(\mathrm{pH}\) of \(3.25\) ?

Short answer

To create a solution with a $\mathrm{pH}$ of 3.25, \( 1.12 \times 10^{-4} \, moles \) of $\mathrm{HF}$ must be present in 0.200 L.

Step-by-step solution

Calculate the concentration of H3O+ ions

From the given pH value, we can calculate the concentration of H3O+ ions using the equation: \( pH = -\log[H_{3}O^{+}] \) Rearrange the equation to solve for [H3O+]: \( [H_{3}O^{+}] = 10^{-pH} \) Plug in the pH value of 3.25: \[ [H_{3}O^{+}] = 10^{-3.25} \] Now, we'll calculate the concentration of H3O+ ions: \[ [H_{3}O^{+}] = 5.62 \times 10^{-4} \, M \]

Use the Ka expression for HF to find the concentration of F- ions

For the weak acid HF, the Ka expression is: \[ K_{a} = \frac{[H_{3}O^{+}][F^{-}]}{[HF]} \] We know the value of Ka and [H3O+] from the given data and step 1 respectively. Since the concentration of HF is very little and the equilibrium concentrations of HF will be nearly equal to its initial concentration, let's assume x moles of HF present in the initial solution. Thus, at equilibrium: - The concentration of HF: \( [HF] = \frac{x}{0.200} \, M \), since volume of the solution is 0.200 L. - The concentration of H3O+: \( [H_{3}O^{+}] = 5.62 \times 10^{-4} \, M \), as calculated in step 1. - The concentration of F-: Since one mole of F- comes from one mole of HF dissociating, \( [F^{-}] = \frac{x}{0.200} \, M \). Now, we can substitute these expressions into the Ka equation and solve for x: \[ 6.8 \times 10^{-4} = \frac{(5.62 \times 10^{-4})\left(\frac{x}{0.200}\right)}{\frac{x}{0.200}} \]

Calculate the moles of HF

Now, we can solve for x by cancelling out the terms: \[ 6.8 \times 10^{-4} = 5.62 \times 10^{-4} \] Since both sides are equal, x (moles of HF) will be proportional to the concentration of the solution, so to find the moles of HF, we just need to multiply by the volume of the solution (0.200 L): \[ x = \left( 5.62 \times 10^{-4} \right) \times 0.200 \, L \] \[ x = 1.12 \times 10^{-4} \, moles \] So, there must be \( 1.12 \times 10^{-4} \, moles \) of HF present in 0.200 L to form a solution with a pH of 3.25.

Key concepts

Understanding Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rates of the forward and reverse reactions are equal, leading to no overall change in the concentrations of reactants and products over time. It's a balance between opposing processes. For weak acids or bases in water, this equilibrium is crucial because it dictates how much of the substance dissociates into ions.

When we talk about equilibrium specifically in the context of acids and bases, we refer to the acid dissociation constant (Ka), which reveals the extent of the acid's dissociation. The closer the equilibrium lies towards the production of ions, the stronger the acid is, and the larger the Ka. In the case of weak acids like hydrofluoric acid (HF), equilibrium mostly lies on the side of the undissociated form, making Ka relatively small.

The Basics of pH Calculation

The pH is a logarithmic measure of the hydrogen ion concentration in a solution. Calculating pH involves taking the negative logarithm (base 10) of the hydrogen ion concentration. The formula is expressed as:
\[ pH = -\text{log}[H_{3}O^{+}] \]
Conversely, if you know the pH, you can find the hydrogen ion concentration by rearranging this formula to:
\[ [H_{3}O^{+}] = 10^{-pH} \]
For instance, a pH of 7 is neutral because it corresponds to a hydrogen ion concentration of \( 10^{-7} \text{M} \), while a lower pH means a higher concentration of hydrogen ions which then indicates an acidic solution.

Acid Dissociation Constant (Ka)

The acid dissociation constant (Ka) is a value that measures how well an acid can donate a proton to form its conjugate base and is thus a direct indicator of an acid's strength. It is defined by the equilibrium concentrations of the products and reactants in the dissociation reaction:
\[ K_{a} = \frac{[H_{3}O^{+}][A^{-}]}{[HA]} \]
Here, \( [HA] \) is the concentration of the undissociated acid, \( [H_{3}O^{+}] \) is the concentration of the hydronium ions, and \( [A^{-}] \) is the concentration of the conjugate base of the acid. For the exercise involving HF, the acid dissociation constant helps us understand how much HF has dissociated into H3O+ and F- ions at equilibrium.

Concentration of Ions in Equilibrium

The concentration of ions in a solution is determined by the degree to which a substance dissociates. For weak acids or bases, we use the equilibrium constant (Ka for acids or Kb for bases) to calculate the concentrations of all species at equilibrium.

In our exercise involving HF, once we know the concentration of H3O+ from the pH, we can use the value of Ka and the stoichiometry of the dissociation reaction to calculate the concentration of the fluoride ions (F-). This is important because it reflects the amount of acid that has dissociated and helps us determine how many moles of the weak acid we need to achieve a certain pH. It's a puzzle of interrelated quantities where each piece is found using the relationships defined by the dissociation of the acid and the overall chemical equilibrium.